If $$x$$ co-ordinates of a point $$P$$ of line joining the points $$Q\left( {2,\,2,\,1} \right)$$ and $$R\left( {5,\,2,\, - 2} \right)$$ is $$4$$, then the $$z$$-coordinates of $$P$$ is :
A.
$$ - 2$$
B.
$$ - 1$$
C.
$$1$$
D.
$$2$$
Answer :
$$ - 1$$
Solution :
Suppose $$P$$ divides $$QR$$ in ratio $$\lambda :1$$
Then, co-ordinates of $$P$$ are $$\left( {\frac{{5\lambda + 2}}{{\lambda + 1}},\,\frac{{\lambda + 2}}{{\lambda + 1}},\,\frac{{ - 2\lambda + 1}}{{\lambda + 1}}} \right)$$
It is given that the $$x$$-coordinate of $$P$$ is $$4.$$
i.e., $$\frac{{5\lambda + 2}}{{\lambda + 1}} = 4 \Rightarrow \lambda = 2$$
So, $$z$$-coordinate of $$P$$ is $$\frac{{ - 2\lambda + 1}}{{\lambda + 1}} = \frac{{ - 4 + 1}}{{2 + 1}} = - 1$$
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :